```University of Phoenix
MTH 209 Algebra II
Week 4
The FUN FUN FUN continues!
• If you have 22=4 Then 2 is the root of four
(or square root)
• If you have 23=8 Then 2 is the cube root
of eight.
Definitions
• nth Roots
• If a=bn for a positive integer n, then :
b is the nth root of a
Specifically if a = b2 then b is the square root
And if a=b3 then b is the cube root
More definitions
• If n is even then you have even roots.
The positive even root of a positive even number is
called the principle root
For example, the principle square root of 9 is 3 The
principle fourth root of 16 is 2
• If n is odd, then you find odd roots.
Because 25 = 32, 2 is the fifth root of 32.
Enter…the…
• We use the
• So we can define :
symbol for a root
n
a
where if n is positive and even and a is positive, then
that denotes the principle nth root of a
If n is a positive odd integer, then it denotes the nth
root of a
If n is any positive integer then n
=0
0
a
as “the nth root of a”
• The n is the index of the radical
• The a is the radicand
Example 1 pg 542
• Find the following roots
2 = 25, the answer is 5
• a)
because
5
25
• b)
 27
3
• c)
6
• d)
 4
64
** Ex. 7-22**
because (-3)3=-27 the answer is –3
because 26 =64 the answer is 2
because
4 =2,
 4 = -2
A look into horror!
• What’s up with these?
9
4
 81
6
 64
• What two numbers (or 4 or 6) satisfy these?
• They are imaginary they are NOT real
numbers and will be NOT be dealt with in
section 9.6 – it was removed in this version
of the class.
Roots and Variables
• Definition: Perfect Squares
x2 ,x4 ,x6 ,x8 ,x10, x12 ,…
EASY to deal with
x x
2
x x
4
2
x x
6
3
Cube roots and Variables
• Definition: Perfect Cubes
x3 ,x6 ,x9 ,x12 ,x15, x18 ,…
EASY to deal with with CUBE roots
3
x x
3
3
x x
6
2
3
x x
9
3
Example 2 – Roots of exponents
page 544
• a)
x
• b)
3
• c)
5
22
x
t t
18
11
6
s s
30
** Ex. 23-34**
6
• What if you have
you want to square it?
2 3
and
( 2  3)  ( 2 )  ( 3 )  2  3  6
2
2
2
• The nth root of a product is equal to the
product of the nth roots. Which looks like:
n
ab  a  b
n
n
provided all of these roots are real numbers.
•
Example 3 Using said product
rule page 545
a)
4y  4  y  2 y  2 y
• b)
3y  3  y  3  y  y
8
8
4
4
By a convention of old people, we normally
put the radical on the right…
3
Ex 3c
c)
3
125w  125 w  5 w
2
** Ex. 35-46**
3
3
2
3
2
Example 4 pg 545
12  4  3  4  3  2 3
a)
b)
c)
3
4
54  3 27 2  3 27  3 2  33 2
80  4 16 5  4 16  4 5  24 5
5
d)
64  5 32 2  5 32  5 2  25 2
** Ex 47-60**
Example 5 pg 546
a)
20x3  4x2  5x  4x2  5x  2x 5x
b)
3
40a8  3 8a6  5a2  3 8a6  3 5a2  2a2 3 5a2
c)
4
48a4b11  4 48a4  b11  4 48a4  4 b11  2ab2 4 3b3
d) 5 w7  5 w5  w2  5 w5  5 w2  w5 w2
** Ex 61-74**
• The nth root of a quotient is equal to the quotient
of the nth roots. In symbols it looks like:
n
n
a
a
n
b
b
• Remember, b can’t equal zero! And all of these
roots must be real numbers.
Example 6 pg 547
a)
25
25 5


9
9 3
c)
3
3
3
b
b
b
3

125
125 5
** Ex. 75-86 **
b)
15
15

 3
3
3
d)
21
3
x

6
y
3
3
x
21
7
x
 2
6
y
y
Example 7 pg 547
Simplify radically with prod. & quot. rule
a)
b)
50
25  2 5 2


49
7
49
c)
4
a5

8
b
4
a4  4 a
4
b8
** Ex. 75-86 **
a4 a
 2
b
5
3
x

8
3
x  x
x x

3
2
8
5
3
2
3
2
Example 8 page 548
Domani’
x 5
a)
You just don’t want the
negative_ number
So x  5  0...x  5
b)
c)
3
4
x7
So all x’s are ok!
2 x  6 This time all x  3 are allowed
•
•
•
•
•
•
•
Definitions Q1-Q6
Find the root in numbers Q7-22
Find the root in variables Q23-Q34
Use the product rule to simplify Q35-Q74
Quotient rule Q75-98
The domain Q99-106
Word problems Q107-117
9.2 Rati0nal Exp0nents
• This is the rest of the story!
• Remember how we looked at the spectrum of
powers?
23=8
22=4
21=2
20=1
2-1=1/2
2-2=1/4
Defining it…
• If n is any positive integer then…
1/ n
a
• Provided that
n
a
 a
n
is a real number
9.2 Ex. 1 pg 553
a)
3
35  35
b)
4
xy  ( xy)
c)
d)
1/ 3
1/ 4
1/ 2
 5
1/ 3
 a
5
a
3
**Ex. 7-14**
Example 2 pg 55
Finding the roots **Ex. 15-22**
a) 41/ 2  4  2
b) (8)
1/ 3
 81  3
1/ 4
4
c) 81
d) (9)
1/ 2
e)  9
  8  2
3
1/ 2
  9not _ real
  9  3
The exponent can be anything!
The numerator is the power
The denominator is root
2
3
Another definition…
a
m/n
 (a )  (a )
1/ n m
m 1/ n
or
a
m/n
 ( a)  a
n
m
n
m
And negatives?
Just upside down.
a
m / n

1
a
m/n
or
a
m / n
1
 n m
( a)
Example 3 pg 554
a)
3
a x
2
1
b)
4
2/3
1
3 / 4
 3/ 4  m
3
m
m
** Ex. 23-26**
Example 4 page 554
a)
b)
5
a
2/3
 5
2 / 5
**Ex. 27-30**
3

2
1
5
a
2
Cookbook 1
a
m / n
Reciprocal
Power
Root
Cookbook 2
1. Find the nth root of a
2. Raise your result to the nth power
3. Find the reciprocal
Example 5 page 555
Rational Expressions
a)
272 / 3  (271/ 3 ) 2  32  9
1 1
4  1/ 2 3  3 
(4 )
2
8
1
1
1
3 / 4
81 
 3
1/ 4 3
(81 )
3
27
1
1
1
1
5 / 3
(8)




1/ 3 5
5
((8) )
(2)
 32
32
3 / 2
b)
c)
d)
**Ex. 31-42**
1
Everything at a glance
(remember these from 2x before?)
Example 6 pg 556
Using product and quotient rules
a)
27  27
1/ 6
3/ 4
b)
1/ 2
1/ 6 1/ 2
 27
 27
2/3
9
5
3 / 4 1/ 4
2/ 4
1/ 2
5
5 5  5
1/ 4
5
**Ex 43-50**
Example 7 pg 557
Power to the Exponents! **Ex.51-60**
a)
1/ 2
3
12
1/ 2
10 1/ 2
b)
(3 )
c)
2
 9
3
6
 (3 12)
1/ 2
 36
2
3
1/ 2
6
3
5
 
 
6 1/ 3
 2
 
9
 3
1 / 3
2
3
27
 3  2 
3
2
4
2 1/ 2
(x )
 x
( sam e_ as)
x  x
2
Ex 8 pg 558 – So you use
absolute values with roots…
a)
b)
8
4 1/ 4
(x y )
x

 8
9
1/ 3



**Ex. 61-70**
 x y _ or _ x y
2
3
x

2
2
Ex 9 pg 558 Mixed Bag **Ex. 71-82**
a) x 2 / 3 y 4 / 3  x 6 / 3  x 2
1/ 2
a
1 / 2 1 / 4
1/ 4
a
a
1
/
4
b) a
c)
3
3 1 / 2
(x y )  (x ) ( y )
1/ 2
 x 
 1/ 3 
d)  y 
2
1 / 2
1/ 2 1/ 2
y
  2
 x
1/ 3
1/ 2



x y
y1 / 6

x
1/ 4
3 / 2
x1/ 4
 3/ 2
y
•
•
•
•
•
•
Definitions Q1-Q6
Rational Exponents Q7-42
Using the rules of Exponents Q43-Q60
Simplifying things with letters Q61-Q82
Mixed Bag Q83-126
Word problems Q127-136
subtracting and multiplying
You treat a radical just like you did a variable.
You could add x’s together (2x+3x = 5x)
And y’s together (4y+10y=14y)
52 5 3 5
and
10 2  2 2  12 2
Ex 1 page 563
a)
3 54 5 7 5
b) 4
w  6 w  5 w
c)
3  5  4 3  6 5  3 3  7 5
d)
4
4
3 6x  2 x  6x  x  4 6x  3 x
3
** Ex. 5-16**
3
3
3
3
3
Ex 2 pg 563
Simplifying then combining
8  18  4  2  9  2  2 2  3 2  5 2
a)
note 8  18  26!!!
2 x  4 x  5 18x  x  2 x  2 x  5  9 x  2 x 
3
b)
2
3
2
2
 x 2 x  2 x  15x 2 x  16x 2 x  2 x
c)
3
16x y  54x y  16x y  2 x  54x y  3 2 x 
4
3
3
4
3
3
4
3
3
 2 xy  3 2 x  3 xy  3 2 x   xy  3 2 x
**Ex. 17-32**
3
4
3
the same index pg 564 **33-46**
a) 5 6  4 3  5  4  6  3  20 18  20  3 2  60 2
b)
3a 2  6a  18a 3  9a 2  2a  3a 2a
3
c)
4  3 4  3 16  3 8  3 2  23 2
3
4
d)
2
5
x 4 x
x
4



2
8
16
4
x  4 x x4 x

4
2
16
4
Ex 4 Multiplying radicals pg 565
** Ex. 47-60**
a)
b)
c)
3 2 (4 2  3 )  3 2  4 2  3 2  3  12  2  3 6  24  3 6
3
a (3 a  3 a 2 )  3 a 2  3 a 3  3 a 2  a
2


3  5 3 3  2 5    FOIL   
 2 3 3 3  2 3  2 5  5 3 3  5  2 5 
 18  4 15  3 15  10  8  15
d)
3 

2
x  9  32  2  3 x  9  ( x  9 ) 2 
 96 x 9  x 9  x 6 x 9
One of those special products reminder
We’ve looked a lot at:
(a+b)(a-b) = a2-b2
So if you multiply two things like this with
radicals, just square the first and last and
subtract them!
Example 6 Multiplying
Conjugates pg566 **Ex. 61-70**
a) (2  3 5 )(2  3 5 )  2  (3 5 )  4  45  41
b) ( 3  2 )( 3  2 )  3  2  1
c)
( 2 x  y )( 2 x  y )  2 x  y
2
2
Example 7 Why not even mix the
indices…why the heck not?
Remember am·an=am+n
a) 3 2  4 2  21/ 3  21/ 4  27 /12  12 27  12 128
b) 3 2  3  21/ 3  31/ 2  22 / 6  33 / 6  6 22  6 33  6 22  33  6 108
•
•
•
•
•
•
•
Definitions Q1-Q4
Multiplying Q33-Q60
Conjugates Q61-Q70
Multiplying different indices Q71-78
Everything Q79-110
Word problems Q127-136
9.4 Quotients and Denominator
Problems
Remember:
2 2 2
3
2 2 2 2
3
3
Example 1 pg 570
**Ex 1-8**
Fix the denominator – no radicals!
a)
b)
3
3 5
15



5
5
5 5
3
3 3 4 33 4 33 4
 3 3  3 
3
2
2
2 4
8
Step by Step help
A radical expression of index n is in
simplified form if it has:
1. No perfect nth power as factors of the
2. No fractions inside the radical, and
3. No radicals in the denominator
Example 2 – Simplifying
a)
10
10 6
60
4 15 2 15
15






6
6
6
3
6
6
6
b)
5 3 5 3 5 3 3 3 15 3 3
 3  3 3  3

9
3
9
9 3
27
3
**Ex. 9-18**
Example 3- Doing the same with
letters. **Ex. 19-28**
a)
a
a b
ab



b
b
b b
b)
x3

5
y
c)
3
x3
y5

x2
y x xy x xy
x
x x
x x

 2
 2

 2

3
4
y

y
y
y y y y y
y
y
x 3 x 3 x 3 y 2 3 xy 2 3 xy 2





2
3
3
y 3 y 3 y 3 y
y
y
n
n
a n a
a bn 
b
b
n
Example 4 Dividing w/same
index
a)
10
10
10  5 

 2
5
5
   
b) 3 2  2 3  3 2  3 2  3  3 6  6
2
2 3 2 3 3 23
2
2
10
x
10
x
c) 3 10x 2  3 5 x 
3
3


2x
3
5x
5x
3
**Ex. 29-36**
Ex 5 – Or simplify BEFORE you
divide
4 3
2 3
3  2x
6x
12  72x 



6x
36  2 x 6 2 x 3 2 x  2 x
a)
4
b)
4
16  4 a
4
2
16a  a 


4
a4  4 a 4 a4 a
4
5
** Ex 37-44**
16
expressions
a) 4  12
4  2 3 2(2  3 ) 2  3



4
4
22
2
 6  20  6  2 5  2(3  5 )
b)


 3 5
2
2
2
**Ex. 45-48
Ex 7 – Rationalizing the
denominator using conjugates
a) 2  3
(2  3 ) (4  3 ) 8  6 3  3 11 6 3




13
13
4  3 (4  3 ) (4  3 )
b)
5
5
( 6  2)
30  10



4
6 2
6  2 ( 6  2)
** Ex. 49-58**
All the tricks again…now things
to powers Ex. 8 pg 575
a)
(5 2 )  5 ( 2 )  125 8  125 2 2  250 2
b)
(2 x )  2
c)
(3w 2 w )  3 w ( 2 w )  27w (2w)  54w
d)
(2t 3t )  2 t ( 3t )  8t
3
3
3 4
3
3
4
3
**Ex. 59-70**
3
( x )  16 x  16x
4
3 4
3
12
3 3
3 3 4
3
3
3
34
27t
3
6
4
Stuff
•
•
•
•
•
•
•
Rationalize the Denominator Q1-8
Using Conjugates Q49-Q58
Powers of Powers Q59-70
Everything Q71-Q108
Word problems Q109-110
Quantum Leap again to
Section 9.5 this time
• Now we do the SAME thing but we are
solving equations and working with word
problems with the RADICALS making a
return.
The odd – root Property
(aren’t they ALL odd?)
• Remember… (-2)3 = -8 and 23 = 8
• So the solution of x3= 8 is 2
and the solution of x3 = -8 is –2
• Because there is only one real odd root of
each real number.
(Flash card time)
The Odd-Root Property
• If n is an odd positive integer
xn = k is equivalent to
for any real number k
x nk
Try it on for size
Example 1 page 579
• a)
x  27, x  3 27, x  3
3
• b) x5  32  0, x5  32, x  5 32, x  2
• c) ( x  2)2  24, x  2  3 24, x  2  2 3 3
**Ex. 5-12**
The Even-Root Property
•
•
•
•
•
•
•
•
Oooh spooky.
If you have x2=4 the answer is 2. Right?
Bzzzt!
We know (2)2=4. Great! BUT (-2)2=4 also.
So the solution is x={2,-2}
Another way to write this is x=±2
So in x4=16, x=±4
And in x6=5 is x=± 6 5
Leaving the book for a page
• Remember John’s fractional exponent trick?
• The solution of x2 is
x
• Let’s solve it:
we have
That’s why
x 1
2
x 1
2
2 1/ 2
(x )
1
2
2
 (1)
x 1
x  1
1/ 2
( x 2 )  (1)
x  1
And the last one from the
previous slide was…
• And in x6=5 is x=±
6
5
x6  5
x 5
( x 6 )1/ 6  (5)1/ 6
6
x
1
6
6
6
 (5)
x  5
1
6
1
6
x  5
6
6
x  6 5
The Even root problem
• In short, if the number inside the even root
is positive, you have a + and – answer.
• If it’s zero, the answer it just zero.
• If it’s negative, you have no solution (in this
universe… it is an imaginary number).
Technically the Even Root
Property looks like
• If k > 0, then xn = k is equivalent to x   n k
• If k = = then xn = 0 is equivalent to x=0
• If k<0, then xn =k has no real solution
Example 2 page 580
using the EVEN root property
• a)
x  10, x   10
2
so _ the _ solution _ set _ is  { 10, 10}or{ 10}
• b) w8=0 so w=0
• c) x4= -4 has no real solution
(to the physicists = 2i, to engineers it is 2j )
**Ex. 13-18**
Example 3 page 581
Using this same property
a) (x-3)2=4
x-3=2 or x-3=-2
x=5 or x=1 The solution set is {1,5}
Example 3b
b)
2( x  5) 2  7  0
2( x  5) 2  7
7
2
7
7
x 5 
, or __ x  5  
2
2
( x  5) 2 
multiply _
7
2
7 2
14
14
by



2
2
2 2
4
2
14
14
, or __ x  5 
2
2
10
14
10
14
x 
, or __ x  
2
2
2
2
10  14
10  14
x
, or __ x 
2
2
x  5
Example 3c
• x4-1=80
• x4=81
• x=± 3 81= ±3
• So the solution set is {-3,3}
** Ex 19-28**
Nonequivalent solutions or
Extraneous Solutions
• When you solve an equation by squaring
both sides, you can get answers that DON’T
satisfy the equation you are working with.
• These are extraneous. Throw them out!
Example 4
• Solve:
• a) 2 x  3  5  0,
2x  3  5
( 2 x  3)  5
2
2
2 x  3  25
2 x  28
x  14 _ or _{14}
b)
3

3 x  5  3 x  1,
3
3x  5
 
3
3
x 1
3x  5  x  1
2 x  6
x  3 _ or _{3}

3
3x  18  x
( 3 x  18)  x
2
3x  18  x
2
2
Ex 4c
**Ex. 29-48**
 x 2  3x  18  0
( x  6)( x  3)  0
x  6  0 __ or __ x  3  0
x  6 _ or _ x  3
Checking
3(3)  18  3 ________ 3(6)  18  6
9  3 _______________ 36  6
3  3 __ bad _ answer !____ 6  6 __ good !!
5x 1  x  2  1
5x 1  1  x  2

5x 1
  1 
2
x2

2
5x 1  1  2 x  2  x  2
5x 1  3  x  2 x  2
And sometimes
you need to
square both
sides twice EX5
**Ex. 49-84**
4x  4  2 x  2
2x  2  x  2
 2x  2
2


x2
Checking

2
4 x2  8x  4  x  2
4 x2  9 x  2  0
(4 x  1)( x  2)  0
4 x  1  0 _ or _ x  2  0
1
x  ___ or ___ x  2
4
1
x  ___ or ___ x  2
4
Plug _ it _ in _ to _ 5 x  1  x  2  1
1
1
1
9 1 3
5  1 
2 

  1
4
4
4
4 2 2
5  2 1  2  2  9  4  3  2  1
Example 6 page 584 **Ex. 65-76**
2
3
• a) x  4
3
b)
 
3
x

4
 
 
x 2  64
x  8 ___ or ___  8
Both _ work , so  {8,8}
2
3
( w  1)

2
5
4
5


5
(
w

1)

4






1
( w  1) 2 
1024
2

5
w 1  
1
1024
1
1
__ or __ w  1  
32
32
33
31
w
__ or __ w 
32
32
31 33
Both _ work , so  { , }
32 32
w 1 
Ex 7 page 585
Not all good things have a solution…
•
•
•
•
•
(2t-3)-2/3=-1
[(2t-3)-2/3]-3=(-1)-3
(2t-3)2=-1
Error! We can’t take the square root of this!
There is no solution in this universe.
**Ex 77-78**
Aid in Solving these things
1. In raising each side of an equation to an even
power, we can create equations that give
extraneous solutions. Check em!
2. When applying the even-root property,
remember that there is a positive and a negative
root for any positive real number.
3. For equations with rational exponents, raise each
side to a positive or negative integral power first,
then apply the even- or odd- root property.
(Postive fraction – raise to a positive power;
negative fraction – raise to a negative power.)
And… back to a few
applications…
• The distance formula
• If you have a triangle with points (x1,y2) and
(x2,y2) you can use the Pythagorean theorem
to get the distance
a2+b2=c2 becomes
d  ( x2  x1 )  ( y2  y1 )
2
2
Example 8 page 585
• Looking Figure 9.1 we want to know the
distance from first to third base when the
bases are 90feet apart.
x 2  902  902
x 2  8100  8100
x 2  16, 200
x   16, 200  90 2
Then _ we _ ignore _ the _ negative
Putting out fires in section 9.5
•
•
•
•
•
•
Some definitions Q1-Q4
Solving for two or no answers Q13-28
Solving and checking for extraneous answers Q29-64
Solving Q65-98
Word problems Q99-124
Changing Chapters…
Chapter 10.1-10.3
Putting it all together, factoring to
graphing. The big sum up.
Section 10.1 Factoring and
Completing the Square
• This is MORE of the same. Nothing new
EXCEPT square roots may show up.
go down like castor oil.
Review of Factoring
• ax2+bx+c=0
• Where a,b,c are real and
• a isn’t equal to 0 (or that’s cheating).
Review of the Cookbook
1. Write the equation with 0 on the right hand
side ( stuff=0)
2. Factor the left hand side.
3. Use the zero factor property to set each
factor equal to zero.
4. Solve the simplest equations.
5. Check the answers in the original equation.
Example 1 page 610
Solving a quadratic equation by factoring
**Ex. 5-14**
3 x  4 x  15  0
2
(3 x  5)( x  3)  0
3 x  5  0 ___ or ___ x  3  0
3 x  5 ____ or ___ x  3
5
x
3
 5 
So _ the _ solution _ set _ is _  ,3
 3 
Example 2 pg 611
Review of the Even-Root Property
• This should also go down quickly… since you’ve
done it sooooo much!
• If you solve (a-1)2=9 you get…
We _ know _ x 2  k _ is _ also _ x   k
(a  1) 2  9
a 1   9
So _ a  1  3 _ or _ a  1  3
a  4 _ or _ a  2
{2, 4}
**Ex. 15-24**
Completing the Square
(making polynomials the way
YOU want them)
• Can you make factorable polynomials if you are
only given the first two terms?
two of the things added together that make that
middle term that when multiplied together equal
that last term.
• Or, in other words, (b/2)2 is your last term.
The rule for finding the last
term…
• x2+bx has a last term that makes the entire
polynomial look like:
x2+bx+(b/2)2
Ex 3 pg 612
Raiders of the Lost Term
x  8 x  __
2
• a)
(8 / 2) 2  4  4  16
So _ x  8 x  16
2
• b)
x 2  5 x  __
25
(5 / 2)  ( 5 / 2)( 5 / 2) 
4
25
2
So _ x  5 x 
4
2
Ex 3 continued
4
x  x  __
7
1 4 2
2 2
4
(  )  ( )( ) 
2 7
7 7
49
4
4
2
So _ x  x 
7
49
2
• c)
• d)
**Ex. 25-32**
3
x  x  __
2
1 3 2
3 3
9
(  )  ( )( ) 
2 2
4 4 16
3
9
2
So _ x  x 
2
16
2
Example 4 pg 612
Remember the perfect square trinomials?
• We’re looking for things in the form
a2+2ab+b2=(a+b)2
• a) x2+12x+36 = (x+6)2
• b) y2-7y+49/4 = (y-7/2)2
• c) z2-4/3z + 4/9 = (y-2/3)2
**Ex 33-40**
If a=1 then we can complete the
squares… Example 5 pg 613
• Given x2+6x+5=0
• The perfect square who’s first two terms are x2+6x
is x2+6x+9
• So we just add 9 to both sides to FORCE this to be
a perfect square!
• x2+6x+5+9=0+9
• x2+6x+9=9-5
• (x+3)2=4
• Now we solve it…
Solving Ex 5
( x  3)  4
2
x  3   4  2
x  3  2 _ or _ x  3  2
x  1 _ or _ x  5
**Ex. 41-48**
If the coefficient of a isn’t 1…
• Too bad. To make this work, you have to
MAKE it = 1!!
• So divide both sides in their entirety by
whatever is before the a
• For example if you have 2x2+4x+10=8
• Then divide EVERYTHING by 2
• Making it x2+2x+5=4 then work on…
The cookbook for these critters:
Completing the Squares
The coefficient of x2 must be 1
Get only the x2 and x terms alone on the RHS
Add to each side ½ the coefficient of x
Factor the left hand side as the square of a
binomial
5. Apply the even root property (plus or minus the
square root of the remaining number)
6. Solve for x
7. Simplify
1.
2.
3.
4.
Example 6 pg 614
a isn’t 1
** Ex.49-50**
• 2x2+3x-2=0
2 x 2  3x  2 0

2
2
3
2
x  x 1  0
2
3
x 2  x __  1
2
3
9
9
2
x  x   1
2
16
16
2
3
25

x  
4  16

3
25
x 
4
16
3 5
3
5
x   _ or _ x   
4 4
4
4
2 1
8
x   _ or _ x    2
4 2
4
x  3x  6  0
2
x  3 x __  6
9
9
2
x  3x   6 
4
4
2
Example 7 pg 615
2
x -3x-6=0
**Ex. 51-60**
2
3  33

x  
2
4

3
33
x 
2
4
3
33
x 
2
4
 3  33 3  33 
3  33
x
_ or _ 
,

2
2 
 2
Now we’ll disguise dishwashing liquid as hand lotion
Ex 8 pg 615 (square first then solve) ** Ex. 61-64**
• Can we deal with square roots in the
problem?
Checking ...
x  3  153  x
 x  3
2


153  x

Root _ x  9?
2
9  3  ? 153  9
x 2  7 x  144  0
12  144 _ good !
Root _ x  16
( x  9)( x  16)  0
16  3  ? 153
x  9  0 _ or _ x  16  0
x  9 _ or _ x  16
13  169  13 _ BAD !
x 2  6 x  9  153  x
Extraneous _ root
1
3
5


x x2 8
The _ LCD _ is  8 x ( x  2)
Example 9 pg 616
LCD then
1
3
5
8 x( x  2)  8 x( x  2)
 8 x( x  2)
complete
x
x2
8
the squares
8 x  16  24 x  5 x  10 x
2
32 x  16  5 x 2  10 x
5 x 2  42 x  16  0
(5 x  2)( x  8)  0
So _ 5 x  2  0 _ or _ x  8  0
2
x  _ or _ x  8
5
2 
Both _ work _ so _  ,8
5 
** Ex. 65-68**
Toying with the dark side…
imaginary solutions!
Example2 10 pg 616 **Ex. 69-78**
x  4 x  12  0
x  4 x ___  12
2
x 2  4 x  4  12  4
( x  2) 2  8
x  2   8
x  2  i 8  2  i 42
x  2  2i 2
Section 10.1 Try a completed
square on for size!
•
•
•
•
•
•
•
•
•
•
Definitions Q1-4
Review – solve by factoring Q5-14
Use the even root property Q15-24
Finish the perfect square trinomial Q25-32
Factor perfect square trinomials Q33-40
Solve by completing the square Q41-58
Potpourri of problems Q59-66
Word Problems Q95-106
• Or… how to get the answer without doing
ANYTHING that is hard as what we have
• The scientific term for it is: Plug’n’Chug
Remember the standard form?
• ax2+bx+c=0
• We can solve for x and always find out what
x (or the x’s) are.
Developing it…
• We’ll just look at it like one would look at
the Grand Canyon.
• You can enjoy it and get into it if you’re in
good shape.
• (where was I going with this slide?)
ax 2  bx  c  0,
ax 2  bx  c 0
 ,
a
a
b
c
x 2  x   0,
a
a
b
c
x2  x   ,
a
a
2
b
b2
 b 
the _ square _ of _ 1/ 2 _ of _ _ is _    2
a
4a
 2a 
b
b2
c b2
x  x 2   2 ,
a
4a
a 4a
2
2
b 
b 2 c 4a

,
x   2 
2
a
4
a
a
4
a


2
b 
b 2  4ac

,
x  
2a 
4a 2

b
b 2  4ac
x

,
2
2a
4a
b
b 2  4ac
x

,
2a
4a 2
b  b 2  4ac
x
2a
• ax2+bx+c=0 where a isn’t 0
b  b  4ac
x
2a
2
Example 1 pg 623
(become the numbers) **Ex. 7-14**
• x2+2x-15=0
• a=1 b=2 c=-15
b  b  4ac
x
2a
2
2  22  4 1 15
x
,
2(1)
2  4  60
x
,
2
2  64
x
,
2
2  8
x
,
2
2  8
2  8
So, x 
, or ,
,
2
2
x  {5,3}
Example 2 pg 623 Only solution
**Ex. 15-20**
• 4x2-12x+9=0
• a=4 b=-12 c=9
b  b  4ac
x
2a
2
12  (12)  4  4  9
x
,
2(4)
2
12  144  144
x
,
2
12  0 12 3
x


8
8 2
Example 3 pg. 624
Two irrational solutions **Ex. 21-26**
•
• a=2 b=16 c=3
2x2+16x+3=0
6  (6) 2  4  2  3
x
,
2(2)
x
b  b  4ac
x
2a
2
x
x
x
6  36  24 6  12

,
4
4
6  4  3 2  3  2 3

,
4
22
2  3  2 3 2( 3  3)

,
22
22
3  3)
2
Example 4 pg 625
Two imaginary solutions (they are in elsewhere)
**Ex. 27-32**
• x2+x+5=0
• a=1 b=1 c=5
b  b  4ac
x
2a
2
1  (1)  4 1  5 1  19
x

,
2(1)
2
2
1  i 19
x
, remember _ 1  i
2
The big picture
• Use the quick reference guide on
PAGE 538 for all the different ways to solve
ax2+bx+c=0
How many solutions?
Look to the discriminate.
• From the earlier examples, you get two
answers when the b2-4ac is positive
• You get one answer when b2-4ac is = 0.
• And no real answers, only imaginary ones,
when b2-4ac is negative.
Example 5 pg. 626 **Ex. 33-48**
a) x2-3x-5=0
b2-4ac = (-3)2-4*1*(-5)=9+20=29 two real ans.
b) x2=3x-9  x2-3x+9 =0
b2-4ac = (-3)2-4*1*9= 9-36=-27 two imag. answers
c) 4x2-12x+9=0
b2-4ac = (-12)2-4*4*9= 144-144=0
One real ans.
Ex 6 - Application pg 626
**Ex. 77-96**
• If the area of a table is 6 sq ft.
And one side is 2 feet shorter than
the other…what are the dimensions?
2  (2) 2  4 1  6 2  28
x

,
2(1)
2
The Setup x(x-2)=6
Or x2-2x-6=0
a=1,b=-2,c=-6
2  28 2  4  7 2  2 7
x


,
2
2
2
x  1 7
Danger _ 1  7 _ is _ negative  toss _ it
So _ x  1  7
The _ sides _ are _ x _ and _ x  2
Side1:1  7, and 7  1
or _ 3.65 feet _ and _ 1.65 feet
Formula
•
•
•
•
•
•
Definitions Q1-Q6
Solve using the formula Q7-32
How many solutions? Q33-48
Solve it the way you want.. Q49-66
Using a calculator Q67-76
Word Problems Q77-106
Equations
We just won’t get this far this class…sorry!
You can email me and work through it if you
want to!
The material below is no longer
part of MTH 209
• Go back, there are dragons ahead!
Functions”
Definition, If Y is determined by a formula
with X in it, we say y is a quadratic function
of x
y=ax2+bx+c
Example 1 Given a number, what’s the
other (more plugging in)
•
•
•
•
•
•
•
a) y=x2-x-6; given (2, ), ( ,0)
The ()’s are (x,y)
y=22-2-6=4-2-6=-4 So the first is (2,-4)
The other one makes us factor
x2-x-6=0
Which is (x-3)(x+2)=0 so x=3 or –2
This one gives us two answers (3,0) and (-2,0)
example 1b
•
•
•
•
s=-16t2+48t+84 given (0, ), ( ,20)
This time it’s (t,s) inside the ()’s
The first is s=-16(0)2+48(0)+84 = 84
It’s ordered pair is (0,84)
• The second is 20 =-16t2+48t+84 
=-16t2+48t+64 =t2-3t+4 =(t-4)(t+1) so t=4 or –1
Giving us (-1,20) and (4,20) as answers.
Graphing. Plug in all values in
the universe for x, and see what y
is  A parabola.
• They look like this!
Example 2 Graphing
2
y=x
• We can go back to the old – try a few
numbers – method.
x
-2
y=x2 4
-1
0
1
2
1
0
1
4
“Can you picture that?”
• It looks like this! With a positive “a” the
“U” shape opens upward.
Domain and Range
•
•
•
•
The domain is the extent of the graph in X
The range is the extent of the graph in Y
In this graph X =  ,  (domain)
Y is only above and including 0 : [0, )
(range)
Example 3 :
2
y=4-x
• y=4-x2
x
-2
-1 0 1 2
y=4-x2
0
3
4 3 0
Figuring out more quickly…
where is the VERTEX?
• The Vertex is the minimum point (if the
thing opens upward) or maximum point (if
the thing opens downward).
• We can find the vertex by using the front
part of:
2
b  b  4ac
x
2
a
b
Mainly x 
2a
The vertex’s above
• For y=x2
• The vertex is: (0,0)
for y=4-x2
here it is (0,4)
Example 4 Using
• Graph y=-x2-x+2
b (1)
1
x


2a 2(1)
2
y _ there _ is...
y   x2  x  2
1 1
9
 1  1
y       2 
 2
4 4
4
 2  2
 1 9
The _ Vertex _ is _( x, y )    , 
 2 4
2
b
x
2a
Example 4 continued
• Plug them numbers in…
x
-2
-1 -1/2 0 1
y=-x2-x+2
0
2
9/4
2 0
Example 5 Do it some more…
• a) y=x2-2x-8
•
•
•
•
•
b
Using x 
give us x=1, then y=12=2*1-8 = -9
2a
The vertex then is (1,-9)
To find the y-intercept, we can plug in x=0 and find
y=02-2*0-8 = -8 So it’s (0,-8)
To find the x-intercept(s) we can plug in y=0 x2-2x8=0 or (x-4)(x+2)=0 so x=4 or x=-2
Now we have sleuthed out some points and can plot
it…
Example 5a, the graph
Example 5b
• a) s=-16t2+64t
b
t
2a
• Using
give us t=2, then s=-16*22+64(2) = 64
• The vertex then is (2,64) (since it is (t,s))
• To find the s-intercept, we can plug in t=0 and find
s=-16(0)2+64*0 = 0 So it’s (0,0)
• To find the t-intercept(s) we can plug in s=0
-16t2+64t=0 or -16t(t-4)=0 so –16t=0 or t-4=0
t=0 or t=4
the t intercept(s) will be (0,0) and (4,0)
• Now we have sleuthed out some points and can plot it…
Example 5b the picture
Section 10.3
•
•
•
•
•
Definitions Q1-6
Complete the ordered pairs Q7-10
Graph the equations Q11-30
Find the max or min Q31-38
Word Problems Q39-48 and beyond
New 10.4
Section 10.5 “Alas, Jean Luc.
All good things must come to an
end.”
• This time we put much of the earlier material
together to do your favorite thing!
• We’ll graph quadratic INEQUALITIES on the
number line.
• Then you can go run in the beautiful spring air and
feel the joyful burden of learning algebra fall off
Again… it’s just a small step…
• They look like:
• ax2+bx+c > 0
• where a, b,c are real numbers and a isn’t 0
• We can use
, , , 
Example 1
•
>0
• (x+5)(x-2)>0 The product is positive so both may be
negative or both may be positive
x2+3x-10
Value
x+5=0
Where
if x= -5
On the number line
Put a 0 above –5
x+5>0
if x>-5
x+5<0
if x<-5
Put + signs to the right
of –5
Put – signs to the left
of -5
Example 1 continued
Value
x-2=0
Where
if x= 2
On the number line
Put a 0 above 2
x-2>0
if x>2
x-2<0
if x<2
Put + signs to the right
of 2
Put – signs to the left
of 2
Example 2
2x  5x  3
2
2x  5x  3  0
2
(2 x  1)( x  3)  0
•
•
•
•
•
•
•
SOOoooo… one is neg, one is pos. or the opposite.
2x-1=0 if x=1/2
2x-1>0 if x>1/2
2x-1>0 if x<1/2
x+3=0 if x=-3
x+3>0 if x>-3
x+3<0 if x<-3
The cookbook
1. Write the inequality with 0 on the right
2. Factor the quadratic polynomial on the left
3. Make a sign graph showing where each
factor is positive, negative or zero.
4. Use the rules for multiplying signed
numbers to determine which regions
satisfy the original equations.
inequalities
x2
2x  3
2
1
 2,
 0,

x 3
x5
x  4 x 1
• You need an LCD to add fractions
• If you multiply by –1 to solve for x, you must
reverse the inequality sign (but you don’t have to
do anything to the inequality sign if you divide or
multiply by a positive number)
Example 3 A rational inequality
x2
 2,
x 3
x  2 2( x  3)

 0,
x 3
x 3
x  2 2x  6

 0,
x 3 x 3
x  2  2x  6
 0,
x 3
x  8
0
x 3
Ex 3 continued
• x-3=0 if x=3
• x-3>0 if x>3
• x-3<0 if x<3
-x+8=0 if x=8
-x+8 > 0 if x>8
-x+8<0 if x<8
Example 4 Now put a ratio on
both sides…
2
1

 0,
x  4 x 1
2( x  1)
1( x  4)

 0,
( x  4)( x  1) ( x  1)( x  4)
2( x  1)  ( x  4)
 0,
( x  4)( x  1)
2x  2  x  4
 0,
( x  4)( x  1)
x2
0
( x  4)( x  1)
The cookbook for rational
inequalities with a sign graph
1. Rewrite the equation with a 0 on the right hand
side
2. Use only addition and subtraction to get an
equivalent inequality
3. Factor the numerator and denominator if
possible
4. Make a sign graph showing where each factor is
positive, negative and zero.
5. Use the rules for multiplying and dividing
signed numbers to determine the regions that
satisfy the original inequality.
Getting your + and – regions
correct. Using a test point.
• Sometimes you can’t factor the portions of
• Are you stuck?
• Heavens no!
• Why not just use the quadratic equationthen test a few points?
Example 5
• x2-4x-6>0
b  b  4ac
x
,
2a
2
4  162  4(1)(6)
x

2(6)
4  40 4  2 10

 2  10
2
2
Example 5 continued
• We can use these points to divide the line by
•
, 2 

Note:


10 , 2  10, 2  10 , 2  10, 

2  10  5.2, and , 2  10  1.2
• So we’ll choose –2, 0 and 7 as test points.
We plug those into the first equation and see
which are true…
Example 5 finishing it up…
Test Point
Value of x2-4x-6
at test point
-2
0
7
6
-6
15
Sign of x2-4x-6 in
interval of test
point
Positive
Negative
Positive
Test Points Cookbook
1. Rewrite the inequality with 0 on the right
2. Solve the quadratic equation that results from
replacing the inequality symbol with the equals
symbol
3. Locate the solutions to the quadratic equation on
a number line
4. Select a test point in each interval determined by
5. Test each point in the original quadratic
inequality to determine which intervals satisfy
the inequality.
Example 6
• After setting up the problem we have the
equation… P=-x2+80x-1500
• For what x is her profit positive (x =
magazine subscriptions)
-x2+80x-1500>0
x2-80x+1500<0
(x-30)(x-50)<0
The
Final Practice!
Section 10.5
•
•
•
•
Definitions Q1-4
Solve each inequality Q5-16
Do it with rational inequalities Q17-36
Solve each inequality using interval
notation Q37-60
• Word Problems Q61-66
Wrap it up with a final
•
•
•
•
Go forth and multiply.
And factor.
And find roots.
etc…
Example 7 Writing in simplified
form
• Simplify
a)
10
10 6
60
4 15 2 15
15






6
6
6
3
6
6
6
b)
3
3
3
3
3
3
5
5
5 3
15
15
 3  3 3  3

9
3
9
9 3
27
• These square roots (like 2 3 and 5
are irrational numbers. It is customary to
rewrite the fraction with a rational number
in the denominator. That is… rationalize it.
• Remember we can always do this:
2  2  2 since 2  2  4  2
Ex 5 Let’s rationalize some
denominators!
• a) Rationalize:
• b)
3
3 5
15



5
5
5 5
3
3 3 2 3 2 33 4
 3 3 3 
3
2
2
2 2 2
• A radical expression of index n is in
simplified form if it has:
• 1) no perfect nth powers as factors of the
• 2) no fractions inside the radical, and
• 3) no radicals in the denominator
Example 7 Of course, we can
insert variables into this!
• Simplify…
• a)
(look for even things you can work with!)
12x  4x  3  4x  3  2x
6
6
6
3
3
• b)
98 x y  49 x y  2 xy  7 x y
5
9
4
8
2
4
2 xy
Example 8 Working with
• We (traditionally remember) want to get rid
of the
in the denominators…
a
a b
ab


b
b b
b
• a)
• b)
x3

5
y
x3
y5

x2
y4
y x xy x xy
x x
x x
 2
 2

 2

3
y

y
y
y
y
y
y
y
y
x
Of course, why not also complicate
things with cube roots and 4th roots
• a)
3
40x  8x
8
3
6 3
5x  2 x
2
23
5x
2
• b)
4
• c)
3
x y  x y
12
x

y
5
3
3
x

y
12
4
3
3
x
y
3
3
4
4
y
2
y
2
y  x y4 y

3
3
3
xy
y
2
3

3
xy
y
2
```